Primary 6 PSLE Mathematics Semestral Assessment 1 (Mid-Year) Paper 5

Answer Key and Marking Scheme

Subject: Mathematics Primary 6
Topic: Whole Numbers
Paper: SA1 Practice Paper (Version 5)

Section A (2 marks each)

1. 4,060,005

• Working:
  • Millions place: 4
  • Ten-thousands place: 6 (Sixty thousand)
  • Ones place: 5
  • Fill zeros for missing places: 4,060,005.
• Teaching Note: Be careful with place values. "Sixty thousand" means 6 is in the ten-thousands column, not the thousands.

2. 8,500,000

• Working:
  • Identify the digit in the hundred-thousands place: 4.
  • Look at the digit to its right (ten-thousands place): 5.
  • Since $5 \ge 5$, round up.
  • $8,400,000 + 100,000 = 8,500,000$.
• Teaching Note: Rounding rules: 0-4 round down, 5-9 round up.

3. 7,200

• Working:
  • Use the distributive property: $a \times b + a \times c = a \times (b + c)$.
  • $72 \times (25 + 75)$
  • $72 \times 100 = 7,200$.
• Teaching Note: Recognizing common factors simplifies calculation significantly.

4. 7

• Working:
  • $4,567 \div 12$
  • $45 \div 12 = 3$ rem $9$
  • $96 \div 12 = 8$ rem $0$
  • $7 \div 12 = 0$ rem $7$
  • Quotient is 380, Remainder is 7.
• Teaching Note: Always check if the remainder is less than the divisor ($7 < 12$).

5. 80

• Working:
  • Let the other number be $x$.
  • $45 \times x = 3,600$
  • $x = 3,600 \div 45$
  • $x = 80$.
• Teaching Note: Division is the inverse of multiplication.

Section B (2 marks each)

6. 300,450; 304,500; 340,050; 345,000

• Working:
  • Compare digits from left to right.
  • All start with 3.
  • Ten-thousands: 0, 0, 4, 4. So 300,450 and 304,500 are smaller.
  • Compare 300,450 and 304,500: Thousands digit 0 < 4. So 300,450 is smallest.
  • Compare 340,050 and 345,000: Thousands digit 0 < 5. So 340,050 < 345,000.
• Teaching Note: Align numbers vertically to compare place values easily.

7. 10,359

• Working:
  • Smallest 5-digit number: Start with smallest non-zero digit for ten-thousands place $\rightarrow$ 1.
  • Next smallest digits for thousands, hundreds, tens $\rightarrow$ 0, 3, 5.
  • Must be odd: Last digit must be 1, 3, 5, or 9.
  • Remaining digit is 9. If we put 9 at the end, the number is 10,359.
  • Check if smaller odd number exists: If last digit is 5, remaining digits 0,3,9 $\rightarrow$ 10,395 (Larger). If last digit is 3, remaining 0,5,9 $\rightarrow$ 10,593 (Larger). If last digit is 1, cannot use 1 again.
  • Smallest is 10,359.
• Teaching Note: "Without repetition" is key. To make a number smallest, put smaller digits at higher place values. To make it odd, the unit digit must be odd.

8. 121

• Working:
  • Order of operations: Brackets first.
  • Inner bracket: $45 \div 5 = 9$.
  • Outer bracket: $20 + 9 = 29$.
  • Subtraction: $150 - 29 = 121$.
• Teaching Note: Follow BODMAS/PEMDAS strictly.

9. 36,250

• Working:
  • February 2024 is a leap year (2024 is divisible by 4).
  • Days in Feb 2024 = 29.
  • Total toys = $1,250 \times 29$.
  • $1,250 \times 30 = 37,500$.
  • $37,500 - 1,250 = 36,250$.
• Teaching Note: Know the number of days in each month and leap year rules.

10. $3,356

• Working:
  • Total spent = $1,299 + 345 = 1,644$.
  • Remaining = $5,000 - 1,644$.
  • $5,000 - 1,600 = 3,400$.
  • $3,400 - 44 = 3,356$.
• Teaching Note: Can also subtract sequentially: $5,000 - 1,299 = 3,701$; $3,701 - 345 = 3,356$.

11. 32

• Working:
  • $8 \times (n - 12) = 160$
  • Divide both sides by 8: $n - 12 = 20$.
  • Add 12 to both sides: $n = 20 + 12 = 32$.
• Teaching Note: Reverse the operations to solve for the unknown.

12. 53

• Working:
  • Let the numbers be $n-1, n, n+1$.
  • Sum = $3n = 156$.
  • $n = 156 \div 3 = 52$.
  • The numbers are 51, 52, 53.
  • Largest is 53.
• Teaching Note: For consecutive numbers, the average is the middle number.

13. 150

• Working:
  • Red = 4 units, Blue = 1 unit.
  • 4 units = 120.
  • 1 unit = $120 \div 4 = 30$ (Blue marbles).
  • Total units = $4 + 1 = 5$ units.
  • Total marbles = $5 \times 30 = 150$.
• Teaching Note: Ratio problems often require finding the value of 1 unit first.

14. $411 \frac{1}{2}$

• Working:
  • $9,876 \div 24$.
  • $98 \div 24 = 4$ rem 2.
  • $27 \div 24 = 1$ rem 3.
  • $36 \div 24 = 1$ rem 12.
  • Quotient 411, Remainder 12.
  • Fraction: $\frac{12}{24} = \frac{1}{2}$.
  • Answer: $411 \frac{1}{2}$.
• Teaching Note: Simplify the remainder fraction.

15. 320

• Working:
  • Friday + Sunday = $1,245 + 1,965 = 3,210$.
  • Saturday = 2,890.
  • Difference = $3,210 - 2,890$.
  • $3,210 - 2,890 = 320$.
  • Wait, the question asks "How many more visitors were there on Saturday than on Friday and Sunday combined?"
  • Saturday (2,890) is less than Combined (3,210).
  • Re-reading question: "How many more visitors were there on Saturday than on Friday and Sunday combined?"
  • This implies Saturday > Combined. But $2,890 < 3,210$.
  • Let's re-read carefully. Usually, this phrasing implies a positive difference. If the question implies Saturday is larger, there is a contradiction in data.
  • However, standard interpretation: Find the difference. Or perhaps I misread the table?
  • Friday 1,245. Sunday 1,965. Sum = 3,210. Saturday 2,890.
  • Perhaps the question meant "How many fewer"? Or "How many more on Combined than Saturday?"
  • Given the phrasing "How many more... on Saturday", and Saturday is smaller, the answer is technically negative or "320 fewer".
  • Correction for Practice Paper Logic: In PSLE, questions are phrased to yield positive integers. Let's assume the question meant "How many more visitors were there on Friday and Sunday combined than on Saturday?"
  • Calculation: $3,210 - 2,890 = 320$.
  • Answer: 320.
• Teaching Note: Always check which quantity is larger before subtracting.

Section C (4 marks each)

16. (a) $1,925 (b)$675

• Working:
  • (a)
    • Sold 450 packets at $4:$450 \times 4 = 1,800$.
    • Remaining packets: $500 - 450 = 50$ packets.
    • Sold 50 packets at $2.50:$50 \times 2.50 = 125$.
    • Total collected: $1,800 + 125 = 1,925$.
  • (b)
    • Cost Price = $1,250.
    • Selling Price = $1,925.
    • Profit = $1,925 - 1,250 = 675$.
• Marking:
  • 1 mark for correct revenue from first batch.
  • 1 mark for correct revenue from second batch.
  • 1 mark for total revenue.
  • 1 mark for correct profit.

17. 233

• Working:
  • Let $N$ be the number of students.
  • $N = 12a + 5$
  • $N = 15b + 8$
  • List numbers between 200 and 250 satisfying condition 1 ($N \div 12$ rem 5):
    • $12 \times 17 = 204 \rightarrow 204 + 5 = 209$.
    • Next: $209 + 12 = 221$.
    • Next: $221 + 12 = 233$.
    • Next: $233 + 12 = 245$.
    • Next: $245 + 12 = 257$ (Out of range).
    • Candidates: 209, 221, 233, 245.
  • Check condition 2 ($N \div 15$ rem 8) for candidates:
    • $209 \div 15 = 13$ rem 14 (No).
    • $221 \div 15 = 14$ rem 11 (No).
    • $233 \div 15 = 15$ rem 8 (Yes).
    • $245 \div 15 = 16$ rem 5 (No).
  • Answer is 233.
• Marking:
  • 1 mark for listing candidates for first condition.
  • 1 mark for checking second condition.
  • 2 marks for correct final answer.

18. (a) $450 (b)$1,950

• Working:
  • This is an arithmetic progression.
  • Jan: 200
  • Feb: 250
  • Mar: 300
  • Apr: 350
  • May: 400
  • Jun: 450
  • (a) June savings = $450.
  • (b) Total = $200 + 250 + 300 + 350 + 400 + 450$.
  • Pairing: $(200+450) + (250+400) + (300+350) = 650 + 650 + 650 = 1,950$.
• Marking:
  • 1 mark for identifying June amount.
  • 1 mark for correct June value.
  • 1 mark for summation method.
  • 1 mark for correct total.

19. (a) $40u$ kg (b) 300

• Working:
  • (a)
    • Let number of large bags = $u$.
    • Mass of large bags = $25 \times u = 25u$ kg.
    • Number of small bags = $3u$.
    • Mass of small bags = $5 \times 3u = 15u$ kg.
    • Total mass expression = $25u + 15u = 40u$ kg.
  • (b)
    • Total mass = 12,000 kg.
    • $40u = 12,000$.
    • $u = 12,000 \div 40$.
    • $u = 300$.
    • Number of large bags is 300.
• Marking:
  • 1 mark for expression for large bags mass.
  • 1 mark for expression for small bags mass.
  • 1 mark for correct combined expression ($40u$).
  • 1 mark for correct value of $u$.

20. (a) 7, 11, 13 (b) 31

• Working:
  • (a)
    • Find prime factors of 1,001.
    • Not divisible by 2, 3 (sum=2), 5.
    • Try 7: $1,001 \div 7 = 143$.
    • Factorize 143. Not divisible by 7 ($143=7 \times 20 + 3$).
    • Try 11: $143 \div 11 = 13$.
    • 13 is a prime number.
    • Prime factors are 7, 11, 13.
  • (b)
    • Sum = $7 + 11 + 13$.
    • $7 + 11 = 18$.
    • $18 + 13 = 31$.
• Marking:
  • 1 mark for finding first factor (7).
  • 1 mark for finding second factor (11).
  • 1 mark for identifying third factor (13).
  • 1 mark for correct sum.